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HACTEHA [7]
3 years ago
12

When you are climbing a hill on a bike with different speeds, why does it seem like you are pedaling a lot to cover a short dist

ance? What happens if you switch to a higher gear in hopes of not having to pedal as much?
Engineering
1 answer:
ololo11 [35]3 years ago
5 0

It seems like your pedaling a lot because it takes more energy and is way slower than a regular road. It you switch to the higher gear it will make you go slower because it is made for going down hill or going high speeds and a lower gear will help you more cause its easier and it will make it go faster.

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A 7-hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, dete
Natali [406]

The maximum volume flow rate of water is determined as 0.029 m³/s.

<h3>Power of the pump</h3>

The power of the pump is watt is calculated as follows;

1 hp = 745.69 W

7 hp = ?

= 7 x 745.69 W

= 5,219.83 W

<h3>Mass flow rate of water</h3>

η = mgh/P

mgh = ηP

m = ηP/gh

m = (0.82 x 5,219.83)/(9.8 x 15)

m = 29.12 kg/s

<h3>Maximum volume rate</h3>

V = m/ρ

where;

  • ρ is density of water = 1000 kg/m³

V = (29.12)/(1000)

V = 0.029 m³/s

Learn more about volume flow rate here: brainly.com/question/21630019

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Norma-Jean [14]

Answer:

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7 0
3 years ago
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The y component of velocity in a steady, incompressible flow field in the xy plane is v = -Bxy3, where B = 0.4 m-3 · s-1, and x
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Answer:

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3 0
3 years ago
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
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